Answer to Question #335217 in General Chemistry for maddy

Question

how many grams of silver iodide, AgI, are produced from 55.3

55.3 g of calcium iodide, CaI₂?

Answer 1

Solution:

The molar mass of CaI₂ is 293.887 g/mol

Therefore,

Moles of CaI₂ = (55.3 g CaI₂) × (1 mol CaI₂ / 293.887 g CaI₂) = 0.1882 mol CaI₂

Balanced chemical equation:

CaI₂ + 2AgNO₃ → 2AgI + Ca(NO₃)₂

According to stoichiometry:

1 mol of CaI₂ produces 2 mol of AgI

Thus, 0.1882 mol of CaI₂ produce:

(0.1882 mol CaI₂) × (2 mol AgI / 1 mol CaI₂) = 0.3764 mol AgI

The molar mass of AgI is 234.77 g/mol

Therefore,

Mass of AgI = (0.3764 mol AgI) × (234.77 g AgI / 1 mol AgI) = 88.367 g AgI = 88.4 g AgI

Mass of AgI = 88.4 g

Answer: 88.4 grams of silver iodide (AgI)

Short form solution:

Balanced chemical equation:

CaI₂ + 2AgNO₃ → 2AgI + Ca(NO₃)₂

According to stoichiometry:

Mass of AgI = (55.3 g CaI₂) × (1 mol CaI₂ / 293.887 g CaI₂) × (2 mol AgI / 1 mol CaI₂) × (234.77 g AgI / 1 mol AgI) = 88.35 g AgI = 88.4 g AgI